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At $300 \mathrm{~K}$, one mole of a gas present in a $10 \mathrm{~L}$ flask exerted a pressure of $2.706 \mathrm{~atm}$.
What is its compressibility factor $(\mathrm{Z})$ ?
(Given $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ).
Options:
What is its compressibility factor $(\mathrm{Z})$ ?
(Given $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ).
Solution:
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Verified Answer
The correct answer is:
$1.1$
$Z=\frac{P V}{R T}=\frac{(2.706)(10)}{(0.082)(300)}=1.1$
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